Simplify and expand the following expression: $ \dfrac{4z + 1}{z - 8}-\dfrac{3z}{4z + 8} $
Answer: In order to subtract expressions, they must have a common denominator. Get both fractions over a common denominator of $(z - 8)(4z + 8)$ Multiply the first term by $\dfrac{4z + 8}{4z + 8}$ $ \begin{align*} \dfrac{4z + 1}{z - 8} \times \dfrac{4z + 8}{4z + 8} & = \dfrac{(4z + 1)(4z + 8)}{(z - 8)(4z + 8)} \\ & = \dfrac{16z^2 + 36z + 8}{(z - 8)(4z + 8)}\end{align*} $ Multiply the second term by $\dfrac{z - 8}{z - 8}$ $ \begin{align*} \dfrac{3z}{4z + 8} \times \dfrac{z - 8}{z - 8} & = \dfrac{(3z)(z - 8)}{(4z + 8)(z - 8)} \\ & = \dfrac{3z^2 - 24z}{(4z + 8)(z - 8)}\end{align*} $ Now we have: $ = \dfrac{16z^2 + 36z + 8}{(z - 8)(4z + 8)} - \dfrac{3z^2 - 24z}{(4z + 8)(z - 8)} $ Now both terms have a common denominator we can subtract the numerators: $ = \dfrac{16z^2 + 36z + 8 - (3z^2 - 24z)}{(z - 8)(4z + 8)} $ $ = \dfrac{16z^2 + 36z + 8 - 3z^2 + 24z}{(z - 8)(4z + 8)} $ $ = \dfrac{13z^2 + 60z + 8}{(z - 8)(4z + 8)}$ Expand the denominator: $ = \dfrac{13z^2 + 60z + 8}{4z^2 - 24z - 64}$